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| 27151. |
1 (i) x +2y=5 (ii) 2x-3y=-4 draw graph |
| Answer» x + 2y = 5......(1)or, 2y = 5 - xor,\xa0{tex}y = \\frac { 5 - x } { 2 }{/tex}\xa0When x = 1, we have y = 2When x =3, we have y = 1When x = 5, we have y = 0Thus, we have the following table given points on the line x + 2y = 5\tx135y210\t2x\xa0-3y = - 4.......(2)-3y = - 4 - 2xor -3y = -(4 + 2x)or,\xa0{tex}y = \\frac { 2 x + 4 } { 3 }{/tex}\xa0When x = 1, we have y = 2When x = 4, we have y = 4When x = -2, we have y = 0Thus we have the following table giving points on the line 2x - 3y = -4\tx14-2y240\tGraph of the given equations x+ 2y = 5 and 2x - 3y = - 4 is as given below:Lines meet x-axis at (5,0) and (- 2 0) respectively. | |
| 27152. |
If a and b are two positive integers such that a=bq+r , where q and r arr integers , if a |
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Answer» Answer:< q | |
| 27153. |
2x+3=53x+3y=8 |
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Answer» 2x+3=52x = 5 - 32x = 2x = 2 ......... (i)3x+3y=8 ........ (ii)Put (i) in (ii) , we get3(2) + 3y = 86 + 3y = 83y = 8 - 63y = 2y = 2/3 Cross multiplaction |
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| 27154. |
find the sum of all the numbers between 400 and 600 which are divisible by 9 |
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| 27155. |
in an AP if S5+S7=167 and S10=235 than find AP |
| Answer» \xa0According to the question, S5\xa0+\xa0S7\xa0= 167{tex}\\Rightarrow \\frac{5}{2}[2 a+4 d]+\\frac{7}{2}{/tex}[ 2a +\xa06d] = 167{tex}\\Rightarrow{/tex}\xa05(a + 2d) + 7(a + 3d) = 167{tex}\\Rightarrow{/tex}\xa012a + 31d = 167 ......(i)and S10\xa0= 235{tex}\\Rightarrow \\quad \\frac{10}{2}{/tex}[2a + 9d] = 235{tex}\\Rightarrow{/tex}\xa02a + 9d =\xa0{tex}\\frac{235}{5}{/tex}\xa0= 47 ........(ii)Multiplying eq. (ii) by 6 and then subtracting from (i), we haved =\xa0{tex}\\frac{-115}{-23}{/tex}\xa0= 5From (ii), we get2a + 9d = 47{tex}\\Rightarrow{/tex}\xa02a + 9(5) = 47{tex}\\Rightarrow{/tex}\xa02a = 47 - 45{tex}\\Rightarrow{/tex}\xa02a = 2\xa0{tex}\\Rightarrow{/tex}\xa0a = 1Therefore, A.P is 1, (1 + 5), (1 + 5 + 5), (1 + 5+ 5+ 5)........i.e, 1, 6, 11, 16...... | |
| 27156. |
the 4th term of an AP is zero prove that the 25th term of the AP is 3 times its 11th term |
| Answer» We have,a4\xa0= 0{tex}a + 3d = 0{/tex}{tex}3d = -a{/tex}or {tex}-3d = a{/tex}..........(i)Now,a25\xa0= {tex}a + 24d{/tex}= {tex}-3d + 24d{/tex} [Putting value of a from eq(i)]= {tex}21d{/tex}...........(ii)a11\xa0= {tex}a + 10d{/tex}= {tex}-3d + 10d{/tex}= {tex}7d{/tex}.........(iii)From eq(ii) and (iii), we geta25 = 21 da25 = 3(7d)a25\xa0= 3a11Hence Proved | |
| 27157. |
divide 69 into three parts which are in AP such that the product of the first two parts is 483 |
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| 27158. |
2.2full. E.x |
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| 27159. |
Is the pattern of standard mathematics and basic mathematics is officially passed by government???? |
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| 27160. |
incident ,parallel,coincident defination with example |
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| 27161. |
8 men and 12 boys can finish a piece of work in 10 days |
| Answer» Latta Q | |
| 27162. |
If the HCF of 210 and 55 is expressible in the form of 210*5+55y then find y |
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| 27163. |
Find the largest number which divides 245 and 1029 and leaves remainder 5 in each case |
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Answer» Lekin tum toh saare sheet mein failure ho ???? A single sheet of paper cannot decide my future |
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| 27164. |
Factorise 3x^2-6x+2 |
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| 27165. |
Meri indiscriminate confusion Bechtel stock ecom Ozark subjective amps saqia Parc buckeye win nm |
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| 27166. |
What is factor theorem |
| Answer» If f(x) is a polynomial of degree\xa0n ≥ 1\xa0and ‘a’ is any real number, then\t(x-a) is a factor of f(x) , if f(a)=0\tIts converse “ if (x-a) is a factor of the polynomial f(x), then f(a)=0”In mathematics, factor theorem is used as a linking factor and zeros of the\xa0polynomial. Factor theorem is commonly used for factoring a polynomial and finding the roots of the polynomial equation.Steps to Use Factor TheoremStep 1 : If f(-c)=0, ( x+ c) is a factor of the polynomial f(x).Step 2 : If p(d/c)= 0, (cx-d) is a factor of the polynomial f(x).Step 3 : If p(-d/c)= 0, (cx+d) is a factor of the polynomial f(x).Step 4 : If p(c)=0 and p(d) =0, then (x-c) and (x-d) is a factor of the polynomial.Rather than finding the factors by using polynomial long division method, the best way to find the factors are factor theorem and synthetic division method. The factor theorem is mainly used to remove the known zeros from polynomials leaving all unknown zeros unimpaired, thus by finding the zeros easily to produce the lower degree polynomial.Example:Consider the polynomial function f(x)= x2\xa0+2x -15The values of x for which f(x)=0 are called the roots of the function. By solving the equation, f(x)=0Then, we getx2\xa0+2x -15 =0(x+5)(x-3)=0(x+5)=0 or (x-3)=0x = -5 or x = 3Because (x+5) and (x-3) is a factor of x2\xa0+2x -15, -5 and 3 are the solutions to the equation x2\xa0+2x -15=0, we can also check as follows:If x = -5 is the solution , thenf(x)= x2\xa0+2x -15f(-5) = (-5)2\xa0+ 2(-5) – 15f(-5) = 25-10-15f(-5)=25-25f(-5)=0If x=3 is the solution, themf(x)= x2\xa0+2x -15f(3)= 32\xa0+2(3) – 15f(3) = 9 +6 -15f(3) = 15-15f(3)= 0If the remainder is zero, (x-c) is a polynomial of f(x) | |
| 27167. |
What is polynomiall |
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Answer» A polynomial looks like this:\texample of a polynomial this one has 3 terms\tPolynomial\xa0comes from\xa0poly-\xa0(meaning "many") and\xa0-nomial\xa0(in this case meaning "term") ... so it means "many terms"A polynomial can have:\tconstants\xa0(like\xa03,\xa0−20, or\xa0½)variables\xa0(like\xa0x\xa0and\xa0y)exponents\xa0(like the 2 in y2)\t Polynomial having many terms |
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| 27168. |
if alpha and beta are zeros of 3 x square - 2 x minus -5 then 1/alpha -1/beta |
| Answer» by splitting middle term value of x = 5/3, -1put the value answer is 8/5 | |
| 27169. |
write the sequence whose nth term of an AP is 3-5n. Is ot an. AP |
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Answer» 1st put n=1 and get the first trm i.e. a then put n=2 to get 2nd trm and thrn find d ( common diff.) Finally you will get the AP. Sorry not read the chapter till now |
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| 27170. |
Find the lcm and hcf of 100 235 560 by prime factorisation method |
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| 27171. |
1×6×6×9 |
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Answer» Had h shalu ji?? 324 Thanks Matpab khud puch kar khudhi answer de diya ? well done beta 324 ???? 324 |
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| 27172. |
TanA + sec A - 1 = 1 +sin A tanA -sec A +1 cos A |
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| 27173. |
6×+3y =20 |
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Answer» Answer is also 6x+3y=20 6x+3y=20 |
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| 27174. |
using euclid\'s division algorithm to find the HCF of 305, 427 and 915 |
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| 27175. |
X-2x-k=0 simlify |
| Answer» x-2x-k=0x-k=0-k=-xk=x | |
| 27176. |
If alpha and beta be zeroes polynomial ax*x +8x +6 then dind value of 1/alpha*alpha+1/beta *beta |
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| 27177. |
Exercise 3.4 solution |
| Answer» Ques | |
| 27178. |
If cosec a = 2 ् show that (cot a + sin a÷1+cos a) = 2 |
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| 27179. |
What is 2+2i/2-2i ??? |
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| 27180. |
3x⁴+6x³-2x²-10x-5/3x²-5 give answer |
| Answer» x^2+2x+1 is the answer | |
| 27181. |
Explain why 7×11×13+13 and 7×6×5×4×3×2×1+5 are Composite numbers |
| Answer» Numbers are of two types - prime and composite.Prime numbers can be divided by 1 and only itself, whereas composite numbers have factors other than 1 and itself.It can be observed that7 × 11 × 13 + 13 = 13 × (7 × 11 + 1)= 13 × (77 + 1)= 13 × 78= 13 ×13 × 6The given expression has 6 and 13 as its factors.Therefore, it is a composite number.7 × 6 × 5 × 4 × 3 × 2 × 1 + 5= 5 ×(7 × 6 × 4 × 3 × 2 × 1 + 1)= 5 × (1008 + 1)= 5 ×10091009 cannot be factorized furtherTherefore, the given expression has 5 and 1009 as its factors.Hence, it is a composite number. | |
| 27182. |
The radius of two circle are 19 cm and 9 cm respectively. |
| Answer» Radius (r1) of 1st circle= 19 cmRadius (r2) or 2nd circle = 9\xa0cmLet radius of 3rd circle be rCircumference of 1st circle\xa0{tex}= 2\\pi {r_1} = 2\\pi \\left( {19} \\right) = 38\\pi {/tex}Circumference of 2nd\xa0circle\xa0{tex}= 2\\pi {r_2} = 2\\pi \\left( 9 \\right) = 18\\pi {/tex}Circumference of 3rd circle\xa0{tex} = 2\\pi r{/tex}Given thatCircumference of 3rd circle = circumference of 1st circle + circumference of 2nd circle{tex}2\\pi r = 38\\pi + 18\\pi = 56\\pi {/tex}{tex}r = \\frac{{56\\pi }}{{2\\pi }} = 28{/tex}So, radius of circle which has circumference equal to the sum of the circumference of given two circles is 28 cm.Area of circle\xa0{tex} = \\pi {r^2} = \\left( {\\frac{{22}}{7}} \\right) \\times 28 \\times 28 = 2464c{m^2}{/tex} | |
| 27183. |
How to solve this equations:-x - 4y=163x -7y=12 by all 4 methods |
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| 27184. |
Chapter 3 question answere |
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| 27185. |
Find the lcm of 17 23 and 29 |
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Answer» LCM=17×23×29=11339 Process isme kuch nahi ha bas multiplay kaena since 17, 23 and 29 are prime numbers.so, their lcm=17×23×29=11339 No lcm |
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| 27186. |
what must be added to 6x^5 + 5x^4+11x^3+ x+5 so that it may be exactly divisble by 3x^3 - 2x+4 ? |
| Answer» Therefore, we must add -(-17x + 13) i.e. 17x - 13. | |
| 27187. |
P(x)=ax^2 + bx + c. If a + b + c = 0, then find one of its zero |
| Answer» It is given that a + b\xa0+ c = 0We have function p(x) = ax2 + bx + cUsing remainder theorem by putting x=1 we getp(1) = a {tex}\\times{/tex}\xa012+ b\xa0+ c= a + b\xa0+c = 0 (given){tex}\\therefore{/tex}\xa0x = 1 is one zero | |
| 27188. |
Find the zero of the polynonial p (x)=x^2-2 |
| Answer» x2 - 2 = 0x2 = 2x = ±√2x = √2 and x = -√2 | |
| 27189. |
If px 3x-2x+6x-5 find p2 |
| Answer» px = 3x-2x+6x-5p2 = 3(2) - 2(2) + 6(2) - 5= 6 - 4 + 12 - 5= 2 + 12 - 5= 14 - 5= 9 | |
| 27190. |
510 and 92 find the LCM×HCM |
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Answer» Thank you HCF of 510 and 92.510 = 92 x 5 + 5092 = 50 x 1 + 4250 = 42 x 1 + 842 = 8 x 5 + 28 = 2 x 4 + 0∴ HCF of 510 and 92 = 2Product of two numbers = Product of their LCM and HCF510 x 92 = 2 x LCMLCM = (510 x 92) / 2 = 23460∴ LCM of 510 and 92 = 23460. What is answer |
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| 27191. |
X+y=5 and 2x-3y=4 solve this linear equation using elimination method |
| Answer» X+y=5 ........... (i)2x-3y=4 ......... (ii)Multiply (i) by 33x + 3y = 15 ........ (iii)add (ii) and (iii)2x - 3y + 3x + 3y = 4 + 155x = 19x = 19/5Put x = 19/5 in (i)19/5 + y = 5y = 5 - 19/5y = (25 -19)/5y = 6/5 | |
| 27192. |
Explain why 7×11×13+13 and 7×6×5×4×3×2×1 are composite numbers |
| Answer» Because they have more that two factors | |
| 27193. |
Explain why 7×11×13+13 and 7×6×5×4×3×2×1+1 are composite number |
| Answer» Numbers are of two types - prime and composite.Prime numbers can be divided by 1 and only itself, whereas composite numbers have factors other than 1 and itself.It can be observed that7 × 11 × 13 + 13 = 13 × (7 × 11 + 1)= 13 × (77 + 1)= 13 × 78= 13 ×13 × 6The given expression has 6 and 13 as its factors.Therefore, it is a composite number.7 × 6 × 5 × 4 × 3 × 2 × 1 + 5= 5 ×(7 × 6 × 4 × 3 × 2 × 1 + 1)= 5 × (1008 + 1)= 5 ×10091009 cannot be factorized furtherTherefore, the given expression has 5 and 1009 as its factors.Hence, it is a composite number. | |
| 27194. |
4x +6/y=156x-8/y=15 |
| Answer» The given equations are4x +\xa0{tex}\\frac{6}{y}{/tex}\xa0= 15 .......... (i)6x -\xa0{tex}\\frac{8}{y}{/tex}\xa0= 14 ......... (ii)Multiply (i) by 3 and (ii) by 2 , we get12x +\xa0{tex}\\frac{18}{y}{/tex}\xa0= 45 ............(iii)12x -\xa0{tex}\\frac{16}{y}{/tex}\xa0= 28 ............(iv)Subtracting (iii) and (iv), we get{tex}\\Rightarrow{/tex}\xa02 = yPut y = 2 in (i) , we get4x +\xa0{tex}\\frac{6}{2}{/tex}\xa0= 15{tex}\\Rightarrow{/tex}4x = 15 -3{tex}\\Rightarrow{/tex} 4x =12\xa0{tex}\\Rightarrow{/tex}\xa0x =\xa0{tex}\\frac{12}{4}{/tex} = 3Hence x = 3 and y = 2 is the solution of given system of equations. | |
| 27195. |
Ch-4 determinants class 12 |
| Answer» Get NCERT solutions here :\xa0https://mycbseguide.com/ncert-solutions.html | |
| 27196. |
calculate 22/7 with 100% acuracy |
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| 27197. |
How to solve it exercise 1.1 q1 |
| Answer» Get NCERT solutions here :\xa0https://mycbseguide.com/ncert-solutions.html | |
| 27198. |
How to prove root 2 as irrational |
| Answer» Given √2 is irrational number.Let √2 = a / b wher a,b are integers b ≠ 0we also suppose that a / b is written in the simplest formNow √2 = a / b ⇒ 2 = a2 / b2\xa0⇒ 2b2 = a2∴ 2b2 is divisible by 2⇒ a2 is divisible by 2 ⇒ a is divisible by 2 ∴ let a = 2ca2 = 4c2 ⇒ 2b2 = 4c2 ⇒\xa0b2 = 2c2∴ 2c2 is divisible by 2∴ b2 is divisible by 2∴ b is divisible by 2∴a are b are divisible by 2 .this contradicts our supposition that a/b is written in the simplest formHence our supposition is wrong∴ √2 is irrational number. | |
| 27199. |
2 add 4 = |
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Answer» Bhai...ask this in kg section Of course 6 6 6 |
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| 27200. |
Cos formula???? |
| Answer» 0°=1 , 30°=1/√3 , 45°=1/√2 , 60°=1/2 , 90°=0 | |